Arc length of parametric curves
Problem 8.94 · medium
Find the length of the curve \( \displaystyle x = e^{t} \cos{\left(t \right)} \), \( \displaystyle y = e^{t} \sin{\left(t \right)} \), \( \displaystyle 0 \le t \le 2 \).
- \[ \left[\begin{matrix}\frac{d}{d t} e^{t} \cos{\left(t \right)}\\\frac{d}{d t} e^{t} \sin{\left(t \right)}\end{matrix}\right] = \left[\begin{matrix}- e^{t} \sin{\left(t \right)} + e^{t} \cos{\left(t \right)}\\e^{t} \sin{\left(t \right)} + e^{t} \cos{\left(t \right)}\end{matrix}\right] \]Velocity components.✓ Proved
- \[ \left(- e^{t} \sin{\left(t \right)} + e^{t} \cos{\left(t \right)}\right)^{2} + \left(e^{t} \sin{\left(t \right)} + e^{t} \cos{\left(t \right)}\right)^{2} = 2 e^{2 t} \](dx/dt)² + (dy/dt)², simplified.✓ Proved
- \[ \int\limits_{0}^{2} \sqrt{2} e^{t}\, dt = - \sqrt{2} \left(1 - e^{2}\right) \]Integrate the speed.✓ Proved
Answer \( - \sqrt{2} \left(1 - e^{2}\right) \approx 9.03549 \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | numerical quadrature of the speed |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly computes the derivatives, simplifies the speed to sqrt(2)e^t, and integrates over the correct bounds. The final expression is algebraically equivalent to the standard form sqrt(2)(e^2 - 1).
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly computes the derivatives, simplifies the speed to sqrt(2)e^t, and integrates over the correct bounds. The final expression is algebraically equivalent to the standard form sqrt(2)(e^2 - 1).gpt-oss:20b: pass 2026-10-05qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly computes the derivatives, simplifies the speed to $\sqrt{2}e^t$, and evaluates the definite integral. The final expression is algebraically equivalent to the standard form $\sqrt{2}(e^2 - 1)$.gpt-oss:20b: pass 2026-10-05
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/parametric_arc_length, checked 2026-10-05 with SymPy 1.14.0.